document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=996415>Question 996415</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1. A father is 34 years old and his son is 12 years old. In how many years will the father be twice as old as his son.<BR>\n" );
document.write( "2. Sam is 18 and Bill is 24. How many years ago was Bill three times as old as her Sam?<BR>\n" );
document.write( "3. John has four times as many marbles as Peter. If he gives Peter 8 marble he will have twice as many marble as Peter. How many did each have originally.<BR>\n" );
document.write( "4. The sum of two numbers is 17, and three times the smaller is 3 more than the bigger, Find the numbers.<BR>\n" );
document.write( "5. If each stroke of a pump removes 20% of the air in the cylinder, what percentage of the air will be left after the third stroke? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #614889 by </B><A HREF=/tutors/aboutme.mpl?userid=ikleyn><B>ikleyn(52786)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ikleyn><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=ikleyn><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=614889\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .<BR>\n" );
document.write( "I'd like to contribute to the question 5) \r<BR>\n" );
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document.write( "\"If each stroke of a pump removes 20% of the air in the cylinder, what percentage of the air will be left after the third stroke?\"<BR>\n" );
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document.write( "More correct formulation is:\r<BR>\n" );
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document.write( "\"If each stroke of a pump removes &nbsp;20%&nbsp; of the &nbsp;<U>current air mass</U>&nbsp; in the cylinder, &nbsp;what percentage of the air will be left after the third stroke?\"\r<BR>\n" );
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document.write( "The answer in the previous response is not correct. &nbsp;If to follow that logic then what percentage of the air will be left after the sixth stroke? Minus &nbsp;20% ??\r<BR>\n" );
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document.write( "The correct analysis is like this:\r<BR>\n" );
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document.write( "1) &nbsp;After the first stroke &nbsp;20% of the initial mass &nbsp;M&nbsp; goes out; &nbsp;and &nbsp;80%, &nbsp;or &nbsp;0.8 &nbsp;of the initial mass, &nbsp;0.8*M, &nbsp;remains.\r<BR>\n" );
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document.write( "2) &nbsp;After the second stroke &nbsp;20% &nbsp;of the remained mass &nbsp;(i.e. &nbsp;0.2*(0.8*M) = 0.16*M) &nbsp;goes out; &nbsp;and &nbsp;0.8*M - 0.16*M = 0.64*M &nbsp;remains.\r<BR>\n" );
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document.write( "3) &nbsp;After the third stroke &nbsp;20% &nbsp;of the remained mass &nbsp;(i.e. 0.2*(0.64*M) = 0.128*M) &nbsp;goes out; &nbsp;and &nbsp;0.64*M - 0.128*M = 0.512*M &nbsp;remains.\r<BR>\n" );
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document.write( "Thus the answer is: &nbsp;After the third stroke &nbsp;0.512 = 51.2% &nbsp;of the initial air mass remains.\r<BR>\n" );
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